jmc

We can write $$c_k=k+\frac{1}{2k+\frac{1}{2k+\frac{1}{2k+\cdots }}}=k+\frac{1}{k+k+\frac{1}{2k+\frac{1}{2k+\cdots }}}=k+\frac{1}{k+c_k}.$$Then $c_k-k=\frac{1}{c_k+k},$ so $c_k^2-k^2=1.$ Hence, $c_k^2=k^2+1.$

Therefore, $$\sum _{k=1}^{11}{c}_k^2=\sum _{k=1}^{11}(k^2+1).$$In general, $$\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6},$$so $$\sum _{k=1}^{11}(k^2+1)=\frac{11\cdot 12\cdot 23}{6}+11=\boxed{517}.$$